# Simplify $\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$ into $\frac{2\sqrt{2x}+\sqrt{2}}{4}$

I am to simplify $$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$$

into $$\frac{2\sqrt{2x}+\sqrt{2}}{4}$$

I am able to get to $$\frac{x+4\sqrt{y}\sqrt{2}}{2}$$ but cannot arrive at the provided solution.

Here is my working:

$$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$$ = $$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{64y}\sqrt{2}}$$ = $$\frac{x+4\sqrt{y}}{\sqrt{2}}$$ = $$\frac{x+4\sqrt{y}}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}$$ = $$\frac{x+4\sqrt{y}\sqrt{2}}{2}$$

I cannot see how to arrive at $$\frac{2\sqrt{2x}+\sqrt{2}}{4}$$?

Screen shot of my online textbooks question and answer in case I've typed it incorrectly: • Are you sure you copied the original equation and the answer correctly? I cannot see how you would have only one $x$ in the original expression, raised to the first power, which would somehow end up under the radical, i.e. $x^{1/2}$, in the answer. – Eevee Trainer Jan 9 at 1:16
• In any event your own attempt is wrong. You cannot cancel the $\sqrt{64}$ like that because there wasn't a $\sqrt{64}$ coefficient for the $y$ term. – Eevee Trainer Jan 9 at 1:17
• Have you tried factoring the $\sqrt{y}$ factor out? – ncmathsadist Jan 9 at 1:18
• Okay, on seeing the edit, small note: $\sqrt{2} x$ means $\sqrt{2} \cdot x$, i.e. the $x$ isn't under the radical. I can see the confusion, I made the same mistake when I was younger. So before I even attempt to address the problem, I just wanted to note that. (If you want less ambiguity, $x \sqrt 2$ is also an acceptable way to write it.) – Eevee Trainer Jan 9 at 1:20
• Yes, but the numerator is a sum. Split it up into its individual fractions, i.e. $$\frac{a+b}{c} = \frac a c + \frac b c$$ and you can see immediately why it doesn't follow. You would have to have the $\sqrt{64}$ in every term of the numerator and every term of the denominator for it to cancel. – Eevee Trainer Jan 9 at 1:22

$$\frac{x \sqrt{64y} + 4 \sqrt{y}}{\sqrt{128 y}}$$ Factor our common factor $$\sqrt{y}$$ from numerator and denominator. $$\frac{\sqrt{y}(x \sqrt{64} + 4)}{\sqrt{y}(\sqrt{128})}$$ Notice $$\sqrt{64} = \sqrt{8^2} = 8$$ and $$\sqrt{128} = \sqrt{64 * 2} = \sqrt{8^2 * 2} = 8\sqrt{2}$$. $$\frac{8x + 4}{8\sqrt{2}}$$ Factor out $$4$$ from numerator and denominator and cancel. $$\frac{2x + 1}{2\sqrt{2}}$$ Multiply numerator and denominator by $$\sqrt{2}$$. $$\frac{2\sqrt{2}x + \sqrt{2}}{4}$$
• $\frac{8x + 4}{8\sqrt{2}} = \frac{4(2x + 1)}{4(2\sqrt{2})} = \frac{2x+1}{2 \sqrt{2}}$ – Klint Qinami Jan 9 at 1:35